17.2 - A Triangular Support

17.2 - A Triangular Support

We now have many of the gory definitions behind us. One of the definitions we learned in particular is that two random variables \(X\) and \(Y\) are independent if and only if:

\(P(X=x, Y=x)=P(X=x)\times P(Y=y)\)

for all \(x\in S_1, y\in S_2\). Otherwise, \(X\) and \(Y\) are said to be dependent. On the previous page, our example comprised two random variables \(X\) and \(Y\), which were deemed to be independent. On this page, we'll explore, by way of another example, two random variables \(X\) and \(Y\), which are deemed to be dependent.

Example 17-2

Consider the following joint probability mass function:


in which the support is \(S=\{(x, y)\}=\{(1, 1), (1, 2), (2,2)\}\). Are the random variables \(X\) and \(Y\) independent?


We are given the joint probability mass function as a formula. We can therefore easily calculate the joint probabilities for each \((x, y)\) in the support \(S\):

when \(x=1\) and \(y=1\): \(f(1,1)=\dfrac{(1)(1)^2}{13}=\dfrac{1}{13}\)

when \(x=1\) and \(y=2\): \(f(1,2)=\dfrac{(1)(2)^2}{13}=\dfrac{4}{13}\)

when \(x=2\) and \(y=2\): \(f(2,2)=\dfrac{(2)(2)^2}{13}=\dfrac{8}{13}\)

Now that we have calculated each of the joint probabilities, we can alternatively present the p.m.f. in tabular form, complete with the marginal p.m.f.s of \(X\) and \(Y\), as:

joint pmf

As an aside, you should note that the joint support \(S\) of \(X\) and \(Y\) is what we call a "triangular support," because, well, it's shaped like a triangle:

joint pmf triangular support

Anyway, perhaps it is easy now to see that \(X\) and \(Y\) are dependent, because, for example:

\(f(1,2)=\dfrac{4}{13} \neq f_X(1)\cdot f_Y(2)=\dfrac{5}{13} \times \dfrac{12}{13}\)

Note though that, in general, any two random variables \(X\) and \(Y\) having a joint triangular support must be dependent because you can always find:

\(f(x)\times f(y)=c\ne0\)

for some non-zero constant \(c\). For example, for the joint p.m.f. above:

\(f_X(2)\times f_Y(1)=\left(\frac{8}{13}\right)\times\left(\frac{1}{13}\right)=\frac{8}{169}\ne 0=f_{X,Y}(2,1)\)

In general, random variables with rectangular support may or may not be independent.

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